A body of mass m is projected with a speed u making an angle of 45...

Consider projection as shown in above diagram.

At the topmost point, velocity of the body will be $ucos45 °$.

Therefore, we can write the required angular momentum as 

$L = \left(m{u}_{H}\right) \times H = \left(mucos\theta\right)\left(\dfrac{{u}^{2}\sin^{2}\theta}{2g}\right)$

$= m{u}^{3} \times \dfrac{1}{4\sqrt{2}g} ⇒ X = 8$